$x^2+y^2-2x = 0$ . is an equation of circle

Here $2g=-2 , 2f=0 $

$$\therefore g=-1\space f=0$$

$\therefore $ center is at $(-g,-f)$ i.e. $(1,0)$ and radius is $\sqrt{g^2+f^2-c}=\sqrt{12}=1$

$$Y=2x$$

It is an equation of line passing from origin. As shown in diagram

$Y^2 = 2x$ is parabola.

The point of intersection of line and parabola is calculated as: $Y^2 = 2x$ and $y = x$

$$\therefore x^2 = 2x $$

$$\therefore x^2-2x=0$$

$$\therefore x(x-2) = 0$$

$$\therefore \space x = 0 or \space x=2 $$

When $x=0 ,y=0 $

When $x=2 , y=2 $

$\therefore$ The vertical strip will slide from 0 to 2 on x-axis.

But the region is divided into two parts as the inner lower limit is different.

Intersection of line and circle

$$X^2+y^2-2x=0$$ and $$y=x$$

$$\therefore x^2+x^2-2x=0 $$

$$\therefore 2x^2-2x=0$$

$$\therefore 2x(x-1)=0$$

$$\therefore \space x=0 or \space x=1 $$

When $x=0 ,y=0$

$$ X=1 , y=1$$

Region (I)

1) Outer limit $x=0 to x=1$

2) Inner limit y

a) Upper limit is equation of parabola i.e. $y^2 = 2x $

$$\therefore y = =\sqrt {2x}$$

b) lower limit is equation of circle i.e.

$$x^2 + y^2 – 2x = 0 $$

$$\therefore y^2 = 2x-x^2$$

$$\therefore y= \sqrt{2x-x^2}$$

Region (II)

1) Outer limit $x=1$ to z

2) Inner limit y.

a) Upper limit is equation of parabola i.e. $y =\sqrt{2x}$

b) Lower limit is equation of line $ y=x $

$\therefore I=\int\limits^1_{x=0}\int\limits_{y=\sqrt{2x-x^2}}^{\sqrt{2x}}xy\space dydx +\int \limits^2_{x=1}\int^{\sqrt {2x}}_{y=x} xy\space dydx\\ =\int\limits_0^1x\Bigg[\dfrac {y^2}2\Bigg]^{\sqrt{2x}}\dfrac {dx}{\sqrt{2x-x^2}}+\int\limits_1^2x\Bigg[\dfrac {y^2}2\Bigg]^{\sqrt{2x}}_xdx\\ =\int\limits_0^1x\Bigg[\dfrac {2x}2-\dfrac {(2x-x^2)}2\Bigg]dx+\int\limits^2_1x\Bigg[\Bigg(\dfrac {2x}2\Bigg)-\dfrac {x^2}2\Bigg]dx\\ =\int\limits_0^1\Bigg(x^2-x^2+\dfrac {x^3}3dx+\int\limits_1^2\Bigg(x^2-\dfrac {x^3}2\Bigg)dx\\ =\Bigg[\dfrac {x^4}8\Bigg]^1_0+\Bigg[\dfrac {x^3}3-\dfrac {x^4}8\Bigg]^2_1\\ =\dfrac 18+\Bigg[\dfrac 83-\dfrac 13-\dfrac {16}8+\dfrac 18\Bigg]\\ =\dfrac 7{12}$